A projectile of wavenumber strikes a nucleus of sharp radius

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · Nuclear Physics · Nuclear reactions

The Nucleus Resolved, Ch Reactions: four-vectors, amplitudes and partial waves

Problem

A projectile of wavenumber kk strikes a nucleus of sharp radius RR, with kR≫1kR \gg 1. ((a)) Starting from the incident flux in channel aa and the outgoing spherical flux in channel bb, derive dσabdΩ=kbka ∣fba(θ)∣2,\frac{\mathrm{d}\sigma_{ab}}{\mathrm{d}\Omega} = \frac{k_b}{k_a}\,|f_{ba}(\theta)|^2, stating where the reduced masses μa\mu_a, μb\mu_b enter and why the factor is a ratio of wavenumbers and not of speeds. ((b)) Specialise to elastic scattering (a=ba=b) and expand f(θ)f(\theta) in partial waves with elasticity parameters ηℓ\eta_\ell, ∣ηℓ∣≤1|\eta_\ell|\le 1. Take the black-disc case ηℓ=0\eta_\ell = 0 for ℓ<kR\ell \lt kR and ηℓ=1\eta_\ell = 1 for ℓ≥kR\ell \ge kR, and show that σel=σreac=πR2\sigma_{\rm el} = \sigma_{\rm reac} = \pi R^2. ((c)) The elastic distribution of part (b) is measured in the laboratory. Using the solid-angle Jacobian dΩcm=(dΩcm/dΩlab) dΩlab\mathrm{d}\Omega_{\rm cm} = (\mathrm{d}\Omega_{\rm cm}/\mathrm{d}\Omega_{\rm lab})\,\mathrm{d}\Omega_{\rm lab}, show that integrating the laboratory differential cross section over the full laboratory solid angle reproduces the same πR2\pi R^2 found in the centre-of-mass frame, i.e. that the total elastic cross section does not depend on which frame it is measured in, however forward-peaked the laboratory distribution itself may be.

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